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My book says that to find the force a current-carrying wire exerts on a moving charge, one uses $B = (\mu_0 / 2\pi)(I/r)$ to find the magnitude of magnetic field around the wire, and then uses that to find $F_B = q v B \sin\theta$, which is the same formula as that for uniform field between the planes of a permanent magnet.

How can the field around the wire ever be uniform since intuitively it weakens as distance increases?

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  • $\begingroup$ Two possibilities here, one you are looking at a question of example that says "Assume a wire runs through a region of uniform magnetic field.." in which case it is not talking about the field from the wire but about some externally imposed field. $\endgroup$ – dmckee --- ex-moderator kitten Jul 28 '15 at 22:51
  • $\begingroup$ Secondly, when we want to make a uniform field we don't use a single, long, straight wire to do it. It won't be long before you are shown geometries that can achieve that to a very good approximation indeed. $\endgroup$ – dmckee --- ex-moderator kitten Jul 28 '15 at 22:52
  • $\begingroup$ @dmckee, no, it's impossible. It's really elementary, won't be that complex $\endgroup$ – most venerable sir Jul 28 '15 at 22:52
  • $\begingroup$ @dmckee, in case i getting it wrong: " the equation for the force of magnetism acting on the charge through a current carrying Wire is the same as when a charge moves in a uniform magnetic field." $\endgroup$ – most venerable sir Jul 28 '15 at 22:57
  • $\begingroup$ Please don't post photos of books, diagrams, or anything else that you can type into the question body. $\endgroup$ – DanielSank Jul 29 '15 at 2:17
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The field around the wire isn't uniform. When you calculate the force on a charge in a magnetic field, you use the value of the field at the point where the particle is. So, $$F_B = qvB\sin\theta$$ is not just for a constant field. If the field varies from position to position in space, then the force the particle feels will also vary. This equation is introduced with a constant field because the math is easier and because the resulting motion is a simple circle.

So, when you calculate the force of a current-carrying wire on a particle, the resulting force is only for that moment in time. If the particle moves towards or away from the wire, the force on the particle will be different. $F_B = qvB\sin\theta$ is always true, but that doesn't mean the force is constant.

Below is a picture of a proton traveling at $10\ m/s$ next to a wire carrying $1\ A$ of current. The proton starts off at the bottom of the picture traveling upwards parallel to the wire in the same direction as the current. 10 m/s proton next to a 1A wire Notice that as the proton gets closer to the wire, the magnetic field increases, so the turn gets tighter. Once the proton turns $180^o$, it starts moving away from the wire, and the turns become wider.

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  • $\begingroup$ If i place a particle between the planes of two permanent magnet, will that charged particle start to move on its own in a circle? $\endgroup$ – most venerable sir Jul 29 '15 at 13:41
  • $\begingroup$ Can you also draw a picture to show how the a electron will circle near the wire? $\endgroup$ – most venerable sir Jul 29 '15 at 13:43
  • $\begingroup$ @user132522 If the charged particle is not moving, the magnets will not exert a force on the particle. I've also updated my answer with a picture. $\endgroup$ – Mark H Jul 29 '15 at 18:49

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